Q) Why didn't we change the units or keep it same?

Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x
m away from the wall.
Then, by Pythagoras theorem, we have:
x² + y² = 25 [Length of the ladder = 5 m]
$y=\sqrt{25-x^2}$
Then, the rate of change of height (y) with respect to time (t) is given by,
$\frac{dy}{dx}=\frac{-x}{\sqrt{25-x^2}}.\frac{dx}{dt}$
It is given that dx/dt = 2 cm/s
$\frac{dy}{dt}=\frac{-2x}{\sqrt{25-x^2}}$
Now, when x = 4 m, we have:
$\frac{dy}{dt}=\frac{-2\times 4}{\sqrt{25-16}}=-\frac{8}{3}$
Hence, the height of the ladder on the wall is decreasing at the rate of 8/3 cm/s.